# STAT 3321 Mod3 HW

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```					                                  Module 3 Homework
Chapter 6 – 6.50 (omit normal probability plot), 6.51abcde (S&P only will be fine)

Chapter 7 – 7.32, 7.49

1. Assume that you own a telephone call center. Over the past several years you have
determined the following information about the length of telephone call you receive. Your
data set of call lengths is normally distributed:

Call length, µ = 150 seconds
2
σ = 225 seconds squared

a. What is the probability that a phone call will last between 150 and 180
seconds?
b. What is the probability that a phone call will last more than 180 seconds?
c. What is the probability that a phone call will last between 125 and 150
seconds?
d. What is the probability that a phone call will last less than 125 seconds?
e. What is the probability that a phone call will last between 145 and 155
seconds?
f. What is the probability that a phone call will last between 140 and 155
seconds?
g. What is the probability that a phone call will last between 160 and 165
seconds?
h. 15% of the phone calls will last less than how many seconds?

2. A manufacturer of tires makes tires with an average life of 36,500 miles (then the tires are
worn out). If the standard deviation is 5,000 miles and the distribution of tire life’s is
normally distributed…
a. What percentage of tires will equal to or exceed 40,000 miles?
b. To provide a warranty such that no more than 10% of the tires would have to be
replaced, at what mileage should the manufacturer set the warranty?

Page 1 of 7
3. Companies that sell stock mutual funds charge their investors expense fees to offset the
mutual fund fees for 137 funds, expressed as a ratio of total assets invested in the fund.
Assume that the distribution of expense ratios is distributed as a normal random variable
with the 137 funds the entire population.

a. Sort the funds by expense ratio.
i. Data, Sort, Expand the Selection, sort ascending by exp. ratio
b. Find the mean and standard deviation via Excel.
c. Create a frequency distribution histogram and graph it.
i. Use a class width of .2, the first class should be .2<=X<.4
d. Find the probability that a stock mutual fund has an expense fee ratio less than
1.5. First, use the Excel function to get the Z value, then use the Excel function
to get the probability (area under the curve).
e. Find the probability that a mutual fund has an expense fee ration less than .7
f. Find the probability that a mutual fund has an expense fee between 1.4 and 2.0.

Excel Functions:

=Average( range )

=Stdevp( range )

=Standardize( X, mean, std. dev.)
returns the Z value

=Normsdist(Z)
returns area under curve to the left

Page 2 of 7
4. The following data represent the number of days absent per year in a population of 6
employees of a small company:
1 3 6 7 7 12
a. compute the population mean and standard deviation
b. select all possible samples of size 2 and set up the sampling distribution of the
mean
c. compute the mean of the sample means, compute the standard error
d. select all possible samples of size 3 and set up the sampling distribution of the
mean
e. compute the mean of the sample means, compute the standard error
f. compare the shape of sampling distributions for b. and d.

5. The diameter of Ping-Pong balls manufactured at a large factory is expected to be
approximately normally distributed with a mean of 1.3 inches and a standard deviation of
.04 inches. What is the probability that a randomly selected Ping-Pong ball will have a
diameter
a. Less than 1.28 inches?
b. Between 1.28 and 1.32 inches?
c. Between what two values (symmetrically distributed around the mean) will 60%
of the Ping-Pong balls fall (in terms of diameter)?

If many random samples of 16 Ping-Pong balls are selected
d. What will be the values of the population mean and standard error of the mean?
e. What distribution will the sample means follow?
f. What proportion of the sample means will be less than 1.28 inches?
g. What proportion of the sample means will be between 1.28 and 1.32 inches?
h. Between what two values (symmetrically distributed around the mean) will 60%
of the sample means be?
i. Which is more likely to occur – an individual ball between 1.28 and 1.32 inches or
a sample mean between 1.28 and 1.32 inches? Explain.

6. A call center has a history of receiving calls that last an average of 150 seconds with a
standard deviation of 15 seconds.
a. What is the probability that a single call is between 150 and 155 seconds?
b. What is the probability that a sample of 35 calls will yield a mean between 150
and 155 seconds?

Page 3 of 7
6.50

Waiting times resemble an exponential distribution, seating times more closely resemble a normal
distribution.

Page 4 of 7
6.51

S&P
a         P(X>0)=P(Z>1.925)                    0.0271
b         P(X>10)=P(Z>2.425)                   0.0077
c         P(X<-50)=P(Z<-0.575)                 0.2826
d         P(X<-60)=P(Z<-1.075)                 0.1412

7.32    a. P(p<.5)=P(Z<-1.2087)=0.1134
b. P(p<.5)=P(Z<-2.7028)=0.0034
c. The standard error is decreased. The sampling distribution of the proportion becomes
more concentrated around the true proportion of 0.56 and, hence, the probability in b. becomes
smaller than in a.

7.49   (a)        24.03        = 10
P(X < 0) = P(Z < -2.403) = 0.0081
(b)     P(10 < X < 20) = P(-1.403 < Z < -0.403) = 0.2632
7.49   (c)     P(X> 10) = P(Z > -1.403) = 0.9197
        10
(d)       X   = 24.03,  X =                =5
n        4
P( X < 0) = P(Z < -4.806) = 0.0000
(e)     P(10 < X < 20) = P(-2.806 < Z < -0.806) = 0.2076
(f)     P( X > 10) = P(Z < -2.806) = 0.9975
(g)     Since the sample mean of return of a sample of treasury bond is distributed closer
than the return of a single treasury bond to the population mean, the probabilities in
(a) and (b) are greater than those in (d) and (e) while the probability in (c) is smaller
than that in (f).

Page 5 of 7

1. µ = 150 seconds, σ = 15 seconds
a. .4772
b. .0228
c. .4525
d. .0475
e. .2586
f. .3779
g. .0927 or 9.27%
h. 134.5 seconds
2. µ = 36,500 miles, σ = 5,000 miles
a. .242 or 24.2%
b. 30,100 miles

3.
Fund             Exp Ratio                                                    X              1.5000
VANGUARD Sm Gr                       0.24                                                    Z              0.3659
FIDELITY SPARTAN Mkt Ind             0.27   Mean                                  1.34       P(X<1.5)       0.6428
AMER CENT Target 2020                0.59   Std Dev Pop                           0.43
PRNCIPL PRESV PSE Tech               0.60                                                    X               0.7000
MAS Mid Cp Gr Inst                   0.62                                     f              Z              -1.5002
LORD ABBETT Affiliated               0.63   .2<=X<.4                                2        P(X<.7)         0.0668
SALOMON Inv Value                    0.63   .4<=X<.6                                1
VANGUARD Select Val                  0.65   .6<=X<.8                               11        X              1.4000
AMER CENT Gl Gold                    0.69   .8<=X<1.0                              13        Z              0.1326
USAA Aggressive Gr                   0.72   1.0<=X<1.2                             24        P(X<1.4)       0.5528
VANGUARD Explorer Fund               0.74   1.2<=X<1.4                             25
PRICE Div Gr                         0.77   1.4<=X<1.6                             23        X              2.0000
GUARDIAN Park Ave A                  0.78   1.6<=X<1.8                             17        Z              1.5322
BABSON Growth                        0.79   1.8<=X<2.0                             10        P(X<2)         0.9373
AMER EXPRESS Stock                   0.82   2.0<=X<2.2                              8
AMER EXPRESS Mutual                  0.83   2.2<=X<2.4                              1        P(X1.4<X<2)    0.3845
Fremont Global                       0.85   2.4<=X<2.6                              2
PIMCO Small cap value Inst           0.85                                         137
SMITH BREEDEN Eq +                   0.88
JANUS Worldwide                      0.89
KEMPER Tech                          0.93                                3-Yr. Returns freq. distribution
FRANKLIN Small Cap Gr A              0.94
Number of Returns

JP MORGAN Sm Co                      0.97                       30
JANUS Enterprise                     0.98                       25
MFS Research A                       0.98                       20
PUTNAM OTC Emerg Gr                  0.98                       15
AMER CENT Vista                      1.00                        5
BRINSON Global equity                1.00                        0
FRANKLIN Dynatech                    1.00
.4

.6

<= 8

.2

.4

.6

.8

.0

.2

.4

.6
0
.

1.

NORTHERN Sm Cp                       1.00
<1

<1

<1

<1

<2

<2

<2

<2
X<

X<

X<

X<

=X

=X

=X

=X

=X

=X

=X

=X
<=

<=

<=

KEMPER Samall cap equity A           1.01
0<

2<

4<

6<

8<

0<

2<

4<
.2

.4

.6

.8

FPA Paramount                        1.03
1.

1.

1.

1.

1.

2.

2.

2.

AMER EXPRESS Discovery               1.05                                                Expense Ratio
FLAG Communications                  1.05
FRANKLIN Global Comm A               1.05
HOTCHKISS and WILEY Small Cap        1.05

Page 6 of 7
4.
X                              X             Sample Mean                                                     X      Sample Mean
1         1,3                                     2                                               1,3,6               3.33
3         1,6                                   3.5                                               1,3,7               3.67
6         1,7                                     4                                               1,3,7               3.67
7         1,7                                     4                                               1,3,12              5.33
7         1,12                                  6.5                                               1,6,7               4.67
12         3,6                                   4.5                                               1,6,7               4.67
3,7                                     5                                               1,6,12              6.33
Mean               6        3,7                                     5                                               3,6,7               5.33
Stdev           3.46        3,12                                  7.5                                               3,6,7               5.33
6,7                                   6.5                                               3,6,12                 7
6 C2             15         6,7                                   6.5                                               6,7,7               6.67
6 C3             20         6,12                                                 9                                  6,7,12              8.33
7,7                                                  7                                  6,7,12              8.33
7,12                                               9.5                                  7,7,12              8.67
7,12                                               9.5                                  1,7,7                  5
1,7,12              6.67
Mean                                                                             6                                  1,7,12              6.67
Std error                                                                     2.45                                  3,7,7               5.67
3,7,12              7.33
3,7,12              7.33
2                 1
Distribution, n=2
3.5               1                                                                                                 Mean                   6
4                 2                                                                                                 Std error           2.00
3.5
4.5               1                    3
5                 2
frequency

2.5
6.5               3                    2
7                 1                  1.5
1
7.5               1                  0.5
9                 1                    0
9.5               2                            2       3.5   4    4.5     5     6.5       7       7.5    9    9.5
TOTAL            15                                                    days absent

3.33              1
3.67              2
Distribution, n=3
4.67              2
5                 1                  3.5
5.33              3                   3
5.67              1                  2.5
frequency

6.33              1                   2
6.67              3                  1.5

7                 1                   1

7.33              2                  0.5

8.33              2                   0
1       2     3    4    5      6     7     8       9     10   11   12
8.67              1                                                     days absent
TOTAL            20

5. a. .3085 somewhat likely
b. .383
c. [1.27, 1.33]
d. 1.3 and .01
e. normal because the original population is normal
f. 0.0228 very unlikely
g. 0.9544
h. [1.29, 1.31]
i. A sample mean. Mean of samples are much more likely to fall near the
population mean.

6. µ = 150 seconds, σ = 15 seconds
a. .1293 or 12.93%
b. .4756 or 47.56% (notice the mean of a sample is much more likely to be in
this interval than an individual observation)

Page 7 of 7

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